Q1: Use suitable identities to find the following products: (i) (x + 4)(x + 10)
Ans: Using the identity (x + a)(x + b) = x2 + (a + b)x + ab, [Here, a = 4 and b = 10] We get, (x + 4)(x + 10) = x2 + (4 + 10)x + (4 x 10) = x2 + 14x + 40
(ii) (x + 8)(x – 10)
Ans: Using the identity, (x+a)(x+b) = x2+(a+b)x+ab [Here, a = 8 and b = (–10)] We get: (x + 8)(x – 10) = x2 + [8 + (–10)]x + [8 x (–10)] = x2 + [8-10]x + [–80] = x2 – 2x – 80
(iii) (3x + 4)(3x – 5)
Ans: Using the identity (x + a)(x + b) = x2 + (a + b)x + ab, [Here, x = 3x, a = 4 and b = −5] we get (3x + 4)(3x – 5) = (3x)2 + [4 + (–5)]3x + [4 x (–5)] = 9x2 + 3x(4–5)–20 = 9x2 – 3x – 20
(iv) (y2 + 3/2) (y2 – 3/2)
Ans: Using the identity (x + y)(x – y) = x2 – y2, [Here, x = y2 and y = 3/2] we get: (y2+3/2)(y2–3/2) = (y2)2–(3/2)2 = y4–9/4
(v) (3 – 2x) (3 + 2x)
Ans: Using (a + b) (a – b) = a2 – b2, putting a = 3 , b = 2x = (3)2 – (2x)2 = 9 – 4x2
Q2: Evaluate the following products without multiplying directly: (i)103 × 107
Ans: (100+3) × (100+7) Using identity, [(x+a)(x+b) = x2+(a+b)x+ab Here, x = 100 a = 3 b = 7 We get, 103×107 = (100+3)×(100+7) = (100)2+(3+7)100+(3×7)) = 10000+1000+21 = 11021
(ii) 95×96
Ans: (100-5)×(100-4) Using identity, [(x-a)(x-b) = x2-(a+b)x+ab Here, x = 100 a = -5 b = -4 We get, 95×96 = (100-5)×(100-4) = (100)2+100(-5+(-4))+(-5×-4) = 10000-900+20 = 9120
(iii)104 × 96
Ans: (100+4)×(100–4) Using identity, [(a+b)(a-b)= a2-b2] Here, a = 100 b = 4 We get, 104×96 = (100+4)×(100–4) = (100)2–(4)2 = 10000–16 = 9984
Q3: Factorise the following using appropriate identities: (i) 9x2 + 6xy + y2
Ans: (3x)2+(2×3x×y)+y2 Using identity, x2+2xy+y2 = (x+y)2 Here, x = 3x y = y 9x2+6xy+y2 = (3x)2+(2×3x×y)+y2 = (3x+y)2 = (3x+y)(3x+y)
(ii) 4y2 – 4y + 1
Ans: 4y2−4y+1 = (2y)2–(2×2y×1)+1 Using identity, x2 – 2xy + y2 = (x – y)2 Here, x = 2y y = 1 4y2−4y+1 = (2y)2–(2×2y×1)+12 = (2y–1)2 = (2y–1)(2y–1)
(iii) x2– y2/100
Ans: x2–y2/100 = x2–(y/10)2 Using identity, x2-y2 = (x-y)(x+y) Here, x = x y = y/10 x2–y2/100 = x2–(y/10)2 = (x–y/10)(x+y/10)
Q4: Expand each of the following, using suitable identities: (i) (x + 2y + 4z)2
Ans: Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx Here, x = x y = 2y z = 4z (x+2y+4z)2 = x22+(2y)2+(4z)2+(2×x×2y)+(2×2y×4z)+(2×4z×x) = x2+4y2+16z2+4xy+16yz+8xz
(ii) (2x – y + z)2
Ans: Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx Here, x = 2x y = −y z = z (2x−y+z)2 = (2x)2+(−y)2+z2+(2×2x×−y)+(2×−y×z)+(2×z×2x) = 4x2+y2+z2–4xy–2yz+4xz
(iii) (–2x + 3y + 2z)2
Ans: Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx Here, x = −2x y = 3y z = 2z (−2x+3y+2z)2 = (−2x)2+(3y)2+(2z)2+(2×−2x×3y)+(2×3y×2z)+(2×2z×−2x) = 4x2+9y2+4z2–12xy+12yz–8xz
(iv) (3a – 7b – c)2
Ans: Using identity (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx Here, x = 3a y = – 7b z = – c (3a –7b– c)2 = (3a)2+(– 7b)2+(– c)2+(2×3a ×– 7b)+(2×– 7b ×– c)+(2×– c ×3a) = 9a2 + 49b2 + c2– 42ab+14bc–6ca
(v) (–2x + 5y – 3z)2
Ans: Using identity, (x+y+z)2= x2+y2+z2+2xy+2yz+2zx Here, x = –2x y = 5y z = – 3z (–2x+5y–3z)2 = (–2x)2+(5y)2+(–3z)2+(2×–2x × 5y)+(2× 5y×– 3z)+(2×–3z ×–2x) = 4x2+25y2 +9z2– 20xy–30yz+12zx
(vi) ((1/4)a-(1/2)b+1)2
Ans: Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx Here, x = (1/4)a y = (-1/2)b z = 1
Q5: Factorise:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz Ans: Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx We can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2 4x2+9y2+16z2+12xy–24yz–16xz = (2x)2+(3y)2+(−4z)2+(2×2x×3y)+(2×3y×−4z) +(2×−4z×2x) = (2x+3y–4z)2 = (2x+3y–4z)(2x+3y–4z)
(ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2 yz – 8xz
Ans: Using identity, (x +y+z)2 = x2+y2+z2+2xy+2yz+2zx We can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2 2x2+y2+8z2–2√2xy+4√2yz–8xz = (-√2x)2+(y)2+(2√2z)2+(2×-√2x×y)+(2×y×2√2z)+(2×2√2×−√2x) = (−√2x+y+2√2z)2 = (−√2x+y+2√2z)(−√2x+y+2√2z)
Q6: Write the following cubes in expanded form: (i) (2x + 1)3
Ans: Using identity,(x+y)3 = x3+y3+3xy(x+y) ((3/2)x+1)3=((3/2)x)3+13+(3×(3/2)x×1)((3/2)x +1)
(iv) (x−(2/3)y)3
Ans: Using identity, (x –y)3 = x3–y3–3xy(x–y)
Q7: Evaluate the following using suitable identities: (i) (99)3
Ans: We can write 99 as 100–1 Using identity, (x –y)3 = x3–y3–3xy(x–y) (99)3 = (100–1)3 = (100)3–13–(3×100×1)(100–1) = 1000000 –1–300(100 – 1) = 1000000–1–30000+300 = 970299 = 970299
(ii) (102)3
Ans: We can write 102 as 100+2 Using identity,(x+y)3 = x3+y3+3xy(x+y) (100+2)3 =(100)3+23+(3×100×2)(100+2) = 1000000 + 8 + 600[100 + 2] = 1000000 + 8 + 60000 + 1200 = 1061208
(iii) (998)3
Ans: We can write 99 as 1000–2 Using identity,(x–y)3 = x3–y3–3xy(x–y) (998)3 =(1000–2)3 =(1000)3–23–(3×1000×2)(1000–2) = 1000000000–8–6000(1000– 2) = 1000000000–8- 6000000+12000 = 994011992
Q8: Factorise each of the following: (i) 8a3 + b3 + 12a2b + 6ab2
Ans: The expression, 8a3+b3+12a2b+6ab2 can be written as (2a)3+b3+3(2a)2b+3(2a)(b)2 8a3+b3+12a2b+6ab2 = (2a)3+b3+3(2a)2b+3(2a)(b)2 = (2a+b)3 = (2a+b)(2a+b)(2a+b) Here, the identity, (x +y)3 = x3+y3+3xy(x+y) is used.
(ii) 8a3 – b3 – 12a2b + 6ab2
Ans: The expression, 8a3–b3−12a2b+6ab2 can be written as (2a)3–b3–3(2a)2b+3(2a)(b)2 8a3–b3−12a2b+6ab2 = (2a)3–b3–3(2a)2b+3(2a)(b)2 = (2a–b)3 = (2a–b)(2a–b)(2a–b) Here, the identity,(x–y)3 = x3–y3–3xy(x–y) is used.
(iii) 27 – 125a3 – 135a + 225a2
Ans: The expression, 27–125a3–135a +225a2 can be written as 33–(5a)3–3(3)2(5a)+3(3)(5a)2 27–125a3–135a+225a2 = 33–(5a)3–3(3)2(5a)+3(3)(5a)2 = (3–5a)3 = (3–5a)(3–5a)(3–5a) Here, the identity, (x–y)3 = x3–y3-3xy(x–y) is used.
(iv) 64a3 – 27b3 – 144 a2b + 108 ab2
Ans: The expression, 64a3–27b3–144a2b+108ab2 can be written as (4a)3–(3b)3 – 3(4a)2(3b)+3(4a)(3b)2 64a3–27b3 – 144a2b+108ab2 = (4a)3–(3b)3–3(4a)2(3b)+3(4a)(3b)2 =(4a–3b)3 =(4a–3b)(4a–3b)(4a–3b) Here, the identity, (x – y)3 = x3 – y3 – 3xy(x – y) is used.
(v) 27p3– (1/216)−(9/2) p2+(1/4)p
Ans: The expression, 27p3–(1/216)−(9/2) p2+(1/4)p can be written as (3p)3–(1/6)3−(9/2) p2+(1/4)p = (3p)3–(1/6)3−3(3p)(1/6)(3p – 1/6) Using (x – y)3 = x3 – y3 – 3xy (x – y) 27p3–(1/216)−(9/2) p2+(1/4)p = (3p)3–(1/6)3−3(3p)(1/6)(3p – 1/6) Taking x = 3p and y = 1/6 = (3p–1/6)3 = (3p–1/6)(3p–1/6)(3p–1/6)
Q10: Factorise each of the following: (i) 27y3+125z3
Ans: The expression, 27y3+125z3 can be written as (3y)3+(5z)3 27y3+125z3 = (3y)3+(5z)3 We know that, x3+y3 = (x+y)(x2–xy+y2) 27y3+125z3 = (3y)3+(5z)3 = (3y+5z)[(3y)2–(3y)(5z)+(5z)2] = (3y+5z)(9y2–15yz+25z2)
(ii) 64m3–343n3
Ans: The expression, 64m3–343n3can be written as (4m)3–(7n)3 64m3–343n3 = (4m)3–(7n)3 We know that, x3–y3 = (x–y)(x2+xy+y2) 64m3–343n3 = (4m)3–(7n)3 = (4m-7n)[(4m)2+(4m)(7n)+(7n)2] = (4m-7n)(16m2+28mn+49n2)
Q11: Factorise 27x3 + y3 + z3 – 9xyz.
Solution: The expression27x3+y3+z3–9xyz can be written as (3x)3+y3+z3–3(3x)(y)(z) 27x3+y3+z3–9xyz = (3x)3+y3+z3–3(3x)(y)(z) We know that, x3+y3+z3–3xyz = (x+y+z)(x2+y2+z2–xy –yz–zx) 27x3+y3+z3–9xyz = (3x)3+y3+z3–3(3x)(y)(z) = (3x+y+z)[(3x)2+y2+z2–3xy–yz–3xz] = (3x+y+z)(9x2+y2+z2–3xy–yz–3xz)
Q12: Verify that x3 + y3 + z3 – 3xyz = (1/2) (x + y + z)[(x – y)2 + (y – z)2 + (z – x)2]
Q13: If x + y + z = 0, show that x3 + y3 + z3 = 3xyz.
Ans: We know that, x3+y3+z3-3xyz = (x +y+z)(x2+y2+z2–xy–yz–xz) Now, according to the question, let (x+y+z) = 0, then, x3+y3+z3 -3xyz = (0)(x2+y2+z2–xy–yz–xz) ⇒ x3+y3+z3–3xyz = 0 ⇒x3+y3+z3 = 3xyz Hence Proved
Q14: Without actually calculating the cubes, find the value of each of the following: (i) (–12)3 + (7)3 + (5)3
Ans: Let a = −12 b = 7 c = 5 We know that if x+y+z = 0, then x3+y3+z3=3xyz. Here, −12+7+5=0 (−12)3+(7)3+(5)3 = 3xyz = 3×-12×7×5 = -1260
(ii) (28)3 + (–15)3 + (–13)3
Ans: Let a = 28 b = −15 c = −13 We know that if x+y+z = 0, then x3+y3+z3 = 3xyz. Here, x+y+z = 28–15–13 = 0 (28)3+(−15)3+(−13)3 = 3xyz = 0+3(28)(−15)(−13) = 16380
Q15: Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given: (i) Area : 25a2–35a+12
Ans: Using the splitting the middle term method, We have to find a number whose sum = -35 and product =25×12=300 We get -15 and -20 as the numbers [-15+-20=-35 and -15×-20=300] 25a2–35a+12 = 25a2–15a−20a+12 = 5a(5a–3)–4(5a–3) = (5a–4)(5a–3) Possible expression for length = 5a–4 Possible expression for breadth = 5a –3
(ii) Area : 35y2+13y–12
Ans: Using the splitting the middle term method, We have to find a number whose sum = 13 and product = 35×-12 = 420 We get -15 and 28 as the numbers [-15+28 = 13 and -15×28=420] 35y2+13y–12 = 35y2–15y+28y–12 = 5y(7y–3)+4(7y–3) = (5y+4)(7y–3) Possible expression for length = (5y+4) Possible expression for breadth = (7y–3)
Q16: What are the possible expressions for the dimensions of the cuboids whose volumes are given below? (i) Volume: 3x2–12x
Ans: 3x2–12x can be written as 3x(x–4) by taking 3x out of both the terms. Possible expression for length = 3 Possible expression for breadth = x Possible expression for height = (x–4)
(ii) Volume: 12ky2+8ky–20k
Ans: 12ky2+8ky–20k can be written as 4k(3y2+2y–5) by taking 4k out of both the terms. 12ky2+8ky–20k = 4k(3y2+2y–5) [Here, 3y2+2y–5 can be written as 3y2+5y–3y–5 using splitting the middle term method.] = 4k(3y2+5y–3y–5) = 4k[y(3y+5)–1(3y+5)] = 4k(3y+5)(y–1) Possible expression for length = 4k Possible expression for breadth = (3y +5) Possible expression for height = (y -1)
